What is the integral of (4x^3-3x+5)/(x^2-2x) ?

The integral of (4x^3-3x+5)/(x^2-2x) is 2x^2+8x-5/2 ln|x|+31/2 ln|x-2|-24+C

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integral of (4x^3-3x+5)/(x^2-2x)

\int \frac{4 x ^{3} - 3 x + 5}{x ^{2} - 2 x} d x

2 x^{2} + 8 x - \frac{5}{2} ln ∣ x ∣ + \frac{31}{2} ln ∣ x - 2 ∣ - 24 + C

Solution steps

\int \frac{4 x ^{3} - 3 x + 5}{x ^{2} - 2 x} d x

Expand

\frac{4 x ^{3} - 3 x + 5}{x ^{2} - 2 x} : \frac{4 x ^{3}}{x ^{2} - 2 x} - \frac{3 x}{x ^{2} - 2 x} + \frac{5}{x ^{2} - 2 x}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{4 x ^{3}}{x ^{2} - 2 x} d x - \int \frac{3 x}{x ^{2} - 2 x} d x + \int \frac{5}{x ^{2} - 2 x} d x

\int \frac{4 x ^{3}}{x ^{2} - 2 x} d x = 4 (\frac{( x - 2 ) ^{2}}{2} + 4 (x - 2) + 4 ln ∣ x - 2 ∣)

\int \frac{3 x}{x ^{2} - 2 x} d x = 3 ln ∣ x - 2 ∣

\int \frac{5}{x ^{2} - 2 x} d x = - 5 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣)

= 4 (\frac{( x - 2 ) ^{2}}{2} + 4 (x - 2) + 4 ln ∣ x - 2 ∣) - 3 ln ∣ x - 2 ∣ - 5 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣)

Simplify

4 (\frac{( x - 2 ) ^{2}}{2} + 4 (x - 2) + 4 ln ∣ x - 2 ∣) - 3 ln ∣ x - 2 ∣ - 5 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣) : 2 x^{2} + 8 x - \frac{5}{2} ln ∣ x ∣ + \frac{31}{2} ln ∣ x - 2 ∣ - 24

= 2 x^{2} + 8 x - \frac{5}{2} ln ∣ x ∣ + \frac{31}{2} ln ∣ x - 2 ∣ - 24

Add a constant to the solution

= 2 x^{2} + 8 x - \frac{5}{2} ln ∣ x ∣ + \frac{31}{2} ln ∣ x - 2 ∣ - 24 + C

\int x^{2} - \frac{1}{( 3 x ) ^{2}} d x

\int \frac{x ^{2} + 8}{x ^{2} - 5 x + 6} d x

\int e^{- a x} cos (w x) d x

\int \frac{1}{2 + cos ( 2 x )} d x

\int \frac{1}{x ln ( a )} d x

What is the integral of (4x^3-3x+5)/(x^2-2x) ?
The integral of (4x^3-3x+5)/(x^2-2x) is 2x^2+8x-5/2 ln|x|+31/2 ln|x-2|-24+C